VDU_Physics_3_005

# ## Solved University-Physics Problems

Physics 3:  Introduction to Electricity & Magnetism

## <<<  Charged Particle in an Electric Field  >>>

Javier Montenegro Joo

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<<<  Charged Particle in an Electric Field   >>>

(2) In the region between two parallel plates separated by a distance d and loaded with opposite electric charges, a horizontal and uniform electric field has been established.  A little ball of mass m and positive electric charge Qo,  is placed at rest at the positive plate.  Sometime t later the ball hits the negative plate. Ignoring the effects of the Gravity, determine (a) the magnitude of the electric field (b) the velocity of the little ball when it hits the negative plate. (c) The kinetic energy of the charged particle when arriving at the negative plate.

Solution

Once the positively charged ball is placed close to the positive plate the former is repelled by the repulsive electrical force of the latter, hence the ball acquires acceleration. According to Newton     F = ma    and  electric field is   E = F/Q   then: (3) In a place where the acceleration of the gravity is    12 m/s2  , an object having an electric charge of   0.000033 C    is placed between two parallel slabs with opposite charges, which setup a uniform vertical electric field of  750 N/C. Determine: (a) The mass of the object if this floats motionless in the field  (b) The magnitude that the electric field should have so that the object acquires an acceleration of    1 m/s2   directed vertically upwards.

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(5) Between two vertical parallel plates loaded with opposite electric charges of the same magnitude, an electric field of magnitude Eo has been established.   A little ball of mass m and positive charge Qo   is placed at rest at a position (xo ,yo) between the plates and, it is displaced due to the electric field and to the gravity. Determine the components (x,y) of the position and (Vx,Vy) of the velocity of the little ball a time t after it has been placed. Show that the trajectory of the little ball inside the plates is a straight line.

Solution

The charged particle is horizontally accelerated by the horizontal electric field between the plates and vertically accelerated by the acceleration of the gravity, which is downwards. Since the particle is simply placed, its initial velocity is zero (Vox = Voy = 0): 